10 prisoners must sample the wine. Bonus points if you worked out a way to ensure than no more than 8 prisoners die.

Number all bottles using binary digits. Assign each prisoner to one of the binary flags. Prisoners must take a sip from each bottle where their binary flag is set.

Here is how you would find one poisoned bottle out of eight total bottles of wine.

          Bottle  | 1 |  2 | 3 | 4 |  5 |  6 | 7 |  8 |
Prisoner A | X |  X | X | X |  X |  X | X |  X |
Prisoner B | X |  X | X | X |  X |  X | X |  X |
Prisoner C | X |  X | X | X |  X |  X | X |  X |
In the above example, if all prisoners die, bottle 8 is bad. If none die, bottle 1 is bad. If A & B dies, bottle 4 is bad.

With ten people there are 1024 unique combinations so you could test up to 1024 bottles of wine.

Each of the ten prisoners will take a small sip from about 500 bottles. Each sip should take no longer than 30 seconds and should be a very small amount. Small sips not only leave more wine for guests. Small sips also avoid death by alcohol poisoning. As long as each prisoner is administered about a millilitre from each bottle, they will only consume the equivalent of about one bottle of wine each.

Each prisoner will have at least a fifty percent chance of living. There is only one binary combination where all prisoners must sip from the wine. If there are ten prisoners then there are ten more combinations where all but one prisoner must sip from the wine. By avoiding these two types of combinations you can ensure no more than 8 prisoners die.

Solution to the problem is as follows:

Move
Time
A & B Cross with torch
2
A Return with torch
1
C & D Cross with torch
10
B Return with torch
2
A & B Cross with torch
2
————————
————–
Total Time
17

Separate any 5 coins on side A and 5 coins on side B.

Lets say,
Side A – TTTHH
Side B – HHHTT

Revrse all coins of anyonen side, lets say side B
So
Side A – TTTHH
Side B – TTTHH

We got all similar facing coin, so simple

Johnny’s mother has a child i.e. Johnny

Other two childs are April and May.

So, third child is JOHNNY

“Incorrectly”

There are 110 squares without a rook.

There are 60 squares of size 1×1.
There are 35 squares of size 2×2.
There are 12 squares of size 3×3.
There are 3 squares of size 4×4.

A total of 60 + 35 + 12 + 3 = 110.

The German has the fish and drinks coffee in the green house, which is fourth on the block.